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UPSC 2026 Maths Optional Paper 2 Q2c — Step-by-Step Solution

20 marks · Section A

Contour integration of real integrals using residues · Complex Analysis · asked 10× in 14 yrs · Read the full method →

Question

Use the method of contour integration to prove

Technique

A rational function of $\sin\theta,\cos\theta$ integrated over a full period $[0,2\pi]$ is the standard signature for the unit-circle substitution $z=e^{i\theta}$: the integral becomes a contour integral of a rational function of $z$ round $|z|=1$, and the residue theorem finishes it. Two things carry the marks here. First, $\sin^{2}\theta$ contributes $z^{2}$ to the denominator while the factor $\dfrac{1}{a+b\cos\theta}$ contributes a $z$ to the numerator and $d\theta$ contributes another $z$ to the denominator — the net singularity at the origin is a double pole, not a simple or a triple one, and the residue there must be computed by differentiation. Second, of the two roots of $bz^{2}+2az+b=0$ exactly one lies inside $|z|=1$, and the hypothesis $a>b>0$ is precisely what guarantees this.

Figure to draw (small, 4 cm, and it earns its space): the unit circle $|z|=1$ with a counter-clockwise arrow; a marked point at the origin labelled ”$z=0$, pole of order $2$”; a point on the negative real axis inside the circle labelled $\alpha=\dfrac{-a+\sqrt{a^{2}-b^{2}}}{b}$, and a point on the negative real axis outside labelled $\beta=\dfrac{-a-\sqrt{a^{2}-b^{2}}}{b}$, with the annotation $\alpha\beta=1$. That one figure states the pole inventory, the orientation, and the inside/outside decision — the three things Step 4 has to justify.

Solution

Throughout, $a>b>0$ are real and $\sqrt{a^{2}-b^{2}}$ denotes the positive square root; note $0<\sqrt{a^{2}-b^{2}}<a$.

Step 1 — The integral is proper (one line, but it is a hypothesis check).

For all real $\theta$, $\cos\theta\ge-1$, so

Hence the denominator never vanishes, the integrand is continuous on $[0,2\pi]$, and $I=\displaystyle\int_{0}^{2\pi}\frac{\sin^{2}\theta}{a+b\cos\theta}\,d\theta$ is an ordinary Riemann integral. (This is the first place $a>b$ is used.)

Step 2 — Substitute $z=e^{i\theta}$ and convert to a contour integral.

Let $z=e^{i\theta}$, $\theta:0\to2\pi$. Then $z$ traverses the unit circle $C:|z|=1$ once, counter-clockwise (positive orientation), and

Therefore

Step 3 — Assemble, and read off the order of the pole at the origin.

Multiply $(2.1)$, $(2.2)$ and $d\theta=\dfrac{dz}{iz}$:

One power of $z$ cancels:

Since $\dfrac{1}{i}=-i$, we have $-\dfrac{1}{2i}=\dfrac{i}{2}$, so

The order of the pole at $z=0$ — do the bookkeeping out loud. Three factors contribute powers of $z$ at the origin: $\sin^{2}\theta$ gives $z^{-2}$, the substitution $d\theta=dz/(iz)$ gives $z^{-1}$, and $\dfrac{1}{a+b\cos\theta}=\dfrac{2z}{bz^{2}+2az+b}$ gives $z^{+1}$. Net: $z^{-2}$. Concretely, in $g$ the numerator satisfies $(z^{2}-1)^{2}\big|_{z=0}=1\neq0$ and the factor $bz^{2}+2az+b$ takes the value $b\neq0$ at $z=0$; hence

(Trap: dropping the numerator $z$ of $(2.2)$ leaves $z^{3}$ downstairs and a spurious third-order pole; writing $\sin^{2}\theta=1-\cos^{2}\theta$ and reducing carelessly can instead leave a simple pole. Either error changes the residue at $0$ and silently destroys the answer, so the cancellation above is written out rather than done mentally.)

Step 4 — The other poles: locate them and decide inside/outside.

Write

Because $a>b>0$ we have $a^{2}-b^{2}>0$, so $\alpha,\beta$ are real and distinct; both are negative, and

(product and sum of the roots of $bz^{2}+2az+b=0$).

Which one is inside $C$? With $d=\sqrt{a^{2}-b^{2}}\in(0,a)$ we have $|\alpha|=\dfrac{a-d}{b}$, and

Both sides are positive ($a-b>0$ by hypothesis, $d>0$), so we may square:

which is true. Hence

By $(4.1)$, $|\beta|=\dfrac{1}{|\alpha|}>1$, so $\beta$ lies outside $C$. In particular no pole lies on $C$, so the residue theorem applies. (This is the second, decisive use of $a>b$: if $a=b$ then $d=0$ and $\alpha=\beta=-1$ sits on the contour.)

So inside $C$ the function $g$ has exactly two poles: a double pole at $z=0$ and a simple pole at $z=\alpha$ (simple because $\alpha\neq\beta$ and $\alpha\neq0$, the latter since $d<a$).

Factorising the quadratic,

Step 5 — Residue at the double pole $z=0$.

For a pole of order $2$, $\ \operatorname*{Res}_{z=0}g=\dfrac{1}{(2-1)!}\displaystyle\lim_{z\to0}\frac{d}{dz}\bigl[z^{2}g(z)\bigr]$. Here

Then $N(0)=1$, $N'(z)=4z(z^{2}-1)$ so $N'(0)=0$; $D(0)=b$, $D'(z)=2bz+2a$ so $D'(0)=2a$. By the quotient rule,

Cross-check by Taylor expansion (the residue is the coefficient of $z^{1}$ in $z^{2}g(z)$):

so $z^{2}g(z)=\dfrac1b-\dfrac{2a}{b^{2}}z+O(z^{2})$, giving the same $-\dfrac{2a}{b^{2}}$. ✓

Step 6 — Residue at the simple pole $z=\alpha$.

From $(4.3)$,

Evaluate the three ingredients with $d=\sqrt{a^{2}-b^{2}}$, so that $\alpha=\dfrac{d-a}{b}$ and $\alpha^{2}=\dfrac{(a-d)^{2}}{b^{2}}$:

(i) $\displaystyle \alpha-\beta=\frac{(-a+d)-(-a-d)}{b}=\frac{2d}{b}$.

(ii) Using $d^{2}=a^{2}-b^{2}$,

hence

(iii) $\displaystyle b\,\alpha^{2}(\alpha-\beta)=b\cdot\frac{(a-d)^{2}}{b^{2}}\cdot\frac{2d}{b}=\frac{2d\,(a-d)^{2}}{b^{2}}$ (non-zero, since $d>0$ and $a\ne d$).

Dividing, the factor $(a-d)^{2}$ cancels:

Step 7 — Apply Cauchy’s Residue Theorem and assemble.

$g$ is analytic on and inside $C$ except at $z=0$ and $z=\alpha$, both interior (Step 4), and has no singularity on $C$. By Cauchy’s residue theorem, for the positively oriented $C$,

Substituting into $(3.1)$ and using $i^{2}=-1$:

That is,

Step 8 — Two sanity checks worth one line each.

(i) Positivity. Since $\sqrt{a^{2}-b^{2}}<a$, the right side is positive — as it must be, the integrand being $\ge0$ with $a+b\cos\theta>0$.

(ii) The $b\to0^{+}$ degeneration. $\sqrt{a^{2}-b^{2}}=a\sqrt{1-b^{2}/a^{2}}=a-\dfrac{b^{2}}{2a}+O(b^{4})$, so

which agrees with the direct evaluation $\displaystyle\int_{0}^{2\pi}\frac{\sin^{2}\theta}{a}\,d\theta=\frac{1}{a}\cdot\pi=\frac{\pi}{a}$. ✓ The apparent $1/b^{2}$ blow-up of the stated formula is therefore removable, not a defect.

Answer

obtained from $I=\dfrac{i}{2}\displaystyle\oint_{|z|=1}\frac{(z^{2}-1)^{2}\,dz}{z^{2}(bz^{2}+2az+b)}$ with $\operatorname*{Res}_{z=0}=-\dfrac{2a}{b^{2}}$ (double pole) and $\operatorname*{Res}_{z=\alpha}=\dfrac{2\sqrt{a^{2}-b^{2}}}{b^{2}}$, $\alpha=\dfrac{-a+\sqrt{a^{2}-b^{2}}}{b}$ being the only root of $bz^{2}+2az+b$ inside $|z|=1$.